This should do the trick, to produce the data frame you asked for, using only base R:
df <- data.frame(cond=c(rep("x", times=length(x)),
rep("y", times=length(y))),
rating=c(x, y))
df
cond rating
1 x 1
2 x 2
3 x 3
4 y 100
5 y 200
6 y 300
However, from your initial description, I'd say that this is perhaps a more likely usecase:
df2 <- data.frame(x, y)
colnames(df2) <- c(x_name, y_name)
df2
cond rating
1 1 100
2 2 200
3 3 300
[edit: moved parentheses in example 1]
While this does not answer the question asked, it answers a related question that many people have had:
x <-c(1,2,3)
y <-c(100,200,300)
x_name <- "cond"
y_name <- "rating"
df <- data.frame(x,y)
names(df) <- c(x_name,y_name)
print(df)
cond rating
1 1 100
2 2 200
3 3 300
You can use expand.grid( ) function.
x <-c(1,2,3)
y <-c(100,200,300)
expand.grid(cond=x,rating=y)
Alt simplification of https://stackoverflow.com/users/1969435/gx1sptdtda above:
cond <-c(1,2,3)
rating <-c(100,200,300)
df <- data.frame(cond, rating)
df
cond rating
1 1 100
2 2 200
3 3 300
df = data.frame(cond=c(rep("x",3),rep("y",3)),rating=c(x,y))
Here's a simple function. It generates a data frame and automatically uses the names of the vectors as values for the first column.
myfunc <- function(a, b, names = NULL) {
setNames(data.frame(c(rep(deparse(substitute(a)), length(a)),
rep(deparse(substitute(b)), length(b))), c(a, b)), names)
}
An example:
x <-c(1,2,3)
y <-c(100,200,300)
x_name <- "cond"
y_name <- "rating"
myfunc(x, y, c(x_name, y_name))
cond rating
1 x 1
2 x 2
3 x 3
4 y 100
5 y 200
6 y 300
Source: Stackoverflow.com